Inductor/Electromagnet used to lift up a magnet? Joules to Newtons or Joules to Gauss?

Question

With interest in using an electromagnet to lift up a magnet.

I dove in. I will state general information. The set up is unknown but I'm interested in the physics/electrical engineering involved in calculating the amount of ______ needed to lift up a mass (magnet with magnetic strength x).

I reviewed "electromagnets". I see an inductor produces a magnetic field, a summation of the contributions of electrons and we get some equations.

Regarding inductors, Energy is .5LI². We can adjust the power/energy/magnetic flux in an inductor by varying the voltage/current/turns, etc.

Also since V = dϕ/dt, maybe we can integrate velocity to find the magnetic flux or something but idk if magnetic flux is relevant to lifting a magnet on the ground.

So I found a link to the an equation expressing the force between two magnets in Newtons homework and exercises - calculate force between two magnets - Physics Stack Exchange

I see magnetic force is measured in Gauss.

I've nearly finished relearning Classical Mechanics, so I'm thinking about using F = ma, and I'm not sure how the inductor/"electromagnet" ties into this nicely.

I don't know how to convert an inductors properties into Gauss, nor express it as a function of distance, and I think you guys can help!

I considered using F = kqq/R or some other attraction law, but still idk how to relate an inductor and it's Joules or other properties to "magnetic strength" I can use in a kinematics analysis.

Answer 1

Your description revolves around magnets and mechanical force which makes the term co-energy very relevant. I'm sure you can find plenty of information regarding that.

Here is a small example on how magnetic force could be calculated. You have a coil with core as well as an object with relatively high permeability \$\mu_r\$. Then the air gaps are the main magnetic resistances in the magnetic circuit.

The magnetic resistance of both air gaps is: $$R_L = \frac{2 d}{A_L \mu_0}$$

The magnetomotive force for \$N\$ turns of the coil is: $$\theta_0 = Ni$$

You can then calculate the magnetic co-energy: $$E_c(i,d) = \frac{1}{2}\psi i = \frac{1}{2}\phi \theta_0 = \frac{\theta_0 ^2}{2 R_{L}} = \frac{N^2 i^2 A_{L} \mu_0}{4 d}$$

From the co-energy, the force can be calculated using partial derivative regarding the distance \$d\$: $$F(i,d) = \frac{\partial E_c}{\partial d} = -\frac{N^2 i^2 A_{L} \mu_0}{4 d^2}$$

_{Source: Lecture Notes: Mechatronics and Electrical Drives}

You should be able to use this as a starting point for more complex arrangements. If you are not familiar with magnetic circuits, that might also be worth researching.